【Codeforces Round 334 (Div 2)A】【模拟 水题】Uncowed Forces CF算分模拟

A. Uncowed Forces

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Kevin Sun has just finished competing in Codeforces Round #334! The round was 120 minutes long and featured five problems with maximum point values of 500, 1000, 1500, 2000, and 2500, respectively. Despite the challenging tasks, Kevin was uncowed and bulldozed through all of them, distinguishing himself from the herd as the best cowmputer scientist in all of Bovinia. Kevin knows his submission time for each problem, the number of wrong submissions that he made on each problem, and his total numbers of successful and unsuccessful hacks. Because Codeforces scoring is complicated, Kevin wants you to write a program to compute his final score.

Codeforces scores are computed as follows: If the maximum point value of a problem is x, and Kevin submitted correctly at minute mbut made w wrong submissions, then his score on that problem is . His total score is equal to the sum of his scores for each problem. In addition, Kevin's total score gets increased by 100 points for each successful hack, but gets decreased by 50 points for each unsuccessful hack.

All arithmetic operations are performed with absolute precision and no rounding. It is guaranteed that Kevin's final score is an integer.

Input

The first line of the input contains five space-separated integers m₁, m₂, m₃, m₄, m₅, where m_i (0 ≤ m_i ≤ 119) is the time of Kevin's last submission for problem i. His last submission is always correct and gets accepted.

The second line contains five space-separated integers w₁, w₂, w₃, w₄, w₅, where w_i (0 ≤ w_i ≤ 10) is Kevin's number of wrong submissions on problem i.

The last line contains two space-separated integers h_s and h_u (0 ≤ h_s, h_u ≤ 20), denoting the Kevin's numbers of successful and unsuccessful hacks, respectively.

Output

Print a single integer, the value of Kevin's final score.

Sample test(s)

input

20 40 60 80 1000 1 2 3 41 0

output

4900

input

119 119 119 119 1190 0 0 0 010 0

output

4930

Note

In the second sample, Kevin takes 119 minutes on all of the problems. Therefore, he gets  of the points on each problem. So his score from solving problems is . Adding in 10·100 = 1000points from hacks, his total score becomes 3930 + 1000 = 4930.

#include<stdio.h>#include<iostream>#include<string.h>#include<string>#include<ctype.h>#include<math.h>#include<set>#include<map>#include<vector>#include<queue>#include<bitset>#include<algorithm>#include<time.h>using namespace std;void fre(){freopen("c://test//input.in","r",stdin);freopen("c://test//output.out","w",stdout);}#define MS(x,y) memset(x,y,sizeof(x))#define MC(x,y) memcpy(x,y,sizeof(x))#define MP(x,y) make_pair(x,y)#define ls o<<1#define rs o<<1|1typedef long long LL;typedef unsigned long long UL;typedef unsigned int UI;template <class T1,class T2>inline void gmax(T1 &a,T2 b){if(b>a)a=b;}template <class T1,class T2>inline void gmin(T1 &a,T2 b){if(b<a)a=b;}const int N=0,M=0,Z=1e9+7,ms63=1061109567;int V(int m,int bas,int w){return max(bas*3/10,bas-bas/250*m-50*w);}int m[7],w[7],h;int main(){int ans=0;for(int i=1;i<=5;++i)scanf("%d",&m[i]);for(int i=1;i<=5;++i)scanf("%d",&w[i]);for(int i=1;i<=5;++i)ans+=V(m[i],i*500,w[i]);scanf("%d",&h);ans+=h*100;scanf("%d",&h);ans-=h*50;printf("%d\n",ans);return 0;}/*【trick&&吐槽】这题我心想着——噢，水题。噢，简单模拟。于是匆匆写完就交，然后也收获了一发wa。原因就是我TwT——我为了节省变量，把所有wa的-50分放到后来才算。然而，这个是要在计算分值的时候才生效，这个-50分并不一定会必然减。TwT 花样作死！【题意】CF算分。分值分别为500 1000 1500 2000 2500每道题的得分公式是max(0.3x,(1-m/250)x-50w)hack成功+100，失败-50【类型】模拟 水题【分析】不改变算分规则的情况下，模拟一发即可。*/